Two trains are moving in opposite directions at 6 0 - - - - - and 9 0 - - - - - Their lengths are 1 - 1 0 - - and 0 - 9 - - respectively- What is the time taken by the faster train to cross the slower train in seconds-

The correct option is A 48 Relative speed = (60+90) km/hr = ( 150 ×5/18)m/sec = ( 125/3) m/sec Distance covered = (1.10+0.9) km = 2 km = 2000 m Required time = ...

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Question

Two trains are moving in opposite directions at 60 km/hr and 90 km/hr. Their lengths are 1.10 km and 0.9 km respectively.
What is the time taken by the faster train to cross the slower train (in seconds)?

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Solution

The correct option is C
48

Relative speed = (60+90) km/hr = $(150 \times \frac{5}{18})$ m/sec = $(\frac{125}{3})$ m/sec

Distance covered = (1.10+0.9) km = 2 km = 2000 m

Required time = $(2000 \times \frac{3}{125})$ sec = 48 sec

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Two trains are moving in opposite directions at 60 km/hr and 90 km/hr. Their lengths are 1.10 km and 0.9 km respectively. What is the time taken by the faster train to cross the slower train (in seconds)?

The correct option is C 48 Relative speed = (60+90) km/hr = (150×518)m/sec = (1253) m/sec Distance covered = (1.10+0.9) km = 2 km = 2000 m Required time = (2000×3125)sec = 48 sec

Two trains are moving in opposite directions at 60 km/hr and 90 km/hr. Their lengths are 1.10 km and 0.9 km respectively.
What is the time taken by the faster train to cross the slower train (in seconds)?

The correct option is C
48

Relative speed = (60+90) km/hr = $(150 \times \frac{5}{18})$ m/sec = $(\frac{125}{3})$ m/sec

Distance covered = (1.10+0.9) km = 2 km = 2000 m

Required time = $(2000 \times \frac{3}{125})$ sec = 48 sec